Top Banner
Unit 3 Unit 3 Digital Modulation Prof A K Ni am Prof A K Nig am
111

Digital Modulation Unit 3

Aug 19, 2014

Download

Engineering

Anil Nigam

Digital Modulation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Digital Modulation Unit 3

Unit 3Unit 3Digital Modulation

Prof A K Ni amProf A K Nigam

Page 2: Digital Modulation Unit 3

Syllabus

• Amplitude Shift Keying (ASK)

• Frequency Shift Keying (FSK), FSK Detection Using PLL

• Binary Phase Shift Keying (PSK)‐ Transmitters, Coherent and non coherent detection, Bit and Baud Rate, Bandwidth and Frequency Spectrum BER and Probability of ErrorFrequency Spectrum. BER and Probability of Error.

• Quadrature Phase Shift Keying (QPSK), QPSK Demodulator, Offset QPSK, Comparison of conventional QPSK and Offset QPSK , BPSK, 8 PSK & 16 PSK

• Quadrature Amplitude Modulation (QAM); 8 QAM & 16 QAM transmitters and receiversQAM transmitters and receivers, 

• Band Width efficiency, 

• Carrier Recovery; Squaring Loop & Costas Loop, y; q g p p,

• Differential PSK, DBPSK transmitter and receiver, 9/18/2013 2Lt Col A K Nigam, ITM University, Gurgaon

Page 3: Digital Modulation Unit 3

Binary Modulation schemesBinary Modulation schemes

• ASKASK

• FSK

S• PSK

M M d l ti h• QPSK

M‐ary Modulation schemesQPSK

• QAM

9/18/2013 3Lt Col A K Nigam, ITM University, Gurgaon

Page 4: Digital Modulation Unit 3

Need For M‐ary Modulation• Mathematically stated, the Shannon limit for information capacity

is

• For a standard telephone circuit with a signal to noise power ratio• For a standard telephone circuit with a signal‐to‐noise power ratioof 1000 (30 dB) and a bandwidth of 2.7 kHz, the Shannon limit forinformation capacity is

I = (3 32)(2700) log10 (1 + 1000) = 26 9 kbpsI = (3.32)(2700) log10 (1 + 1000) = 26.9 kbps• Using multilevel signaling, the Nyquist formulation for channel

capacity is I =2B log2 MTh f hi i 26 9 kb h t h• Thus for achieving 26.9 kbps we have to have

26900 =2×2700× log2 M• This gives log2 M=5 or L=32, thus in order to achieve 26.9 kbps we

have to use multilevel signals

9/18/2013 4Lt Col A K Nigam, ITM University, Gurgaon

Page 5: Digital Modulation Unit 3

Bit and BaudBit and Baud

• Mathematically, baud is the reciprocal of the time of oney, poutput signaling element, and a signaling element mayrepresent several information bits.

• Baud is expressed as baud =1/ts

• In addition, since baud is the encoded rate of change, it alsoequals the bit rate fb divided by the number of bits encodedinto one signaling element. Thus,

9/18/2013 5Lt Col A K Nigam, ITM University, Gurgaon

Page 6: Digital Modulation Unit 3

AMPLITUDE‐SHIFT KEYING• Mathematically, amplitude‐shift keying is

• Modulating signal [vm(t)] is normalized where + 1 V = logic 1Modulating signal [vm(t)] is normalized where + 1 V = logic 1 and ‐1 V = logic 0. 

• Therefore for a logic 1 input, vm(t) = +1 V, Equation reduces to

9/18/2013 6Lt Col A K Nigam, ITM University, Gurgaon

Page 7: Digital Modulation Unit 3

• And for a logic 0 input, vm(t) = ‐1 V, Equation reduces to 

• Thus, the modulated wave is either A cos(wct) or 0.

• Hence, the carrier is either "on“ or "off" which is why amplitude‐shift keying is sometimes referred to as on‐p y goff keying(OOK).

9/18/2013 7Lt Col A K Nigam, ITM University, Gurgaon

Page 8: Digital Modulation Unit 3

ASK (also called Digital amplitude modulation, DAM)

Performance of BASK/OKK/DAM

• As noise affects only the amplitude of the signal, the noise performance of this is not good

9/18/2013 8Lt Col A K Nigam, ITM University, Gurgaon

Page 9: Digital Modulation Unit 3

ASK Modulator

9/18/2013 9Lt Col A K Nigam, ITM University, Gurgaon

Page 10: Digital Modulation Unit 3

Asynchronous DemodulatorAsynchronous Demodulator

Envelope Detector

LPFInput                                                                 Demodulated       output

9/18/2013 10Lt Col A K Nigam, ITM University, Gurgaon

Page 11: Digital Modulation Unit 3

Synchronous ASK Demodulation

[ ]( / 2 )c b cf f f± ±

( / 2 )c bf f±

[ ]( )c b cf f f

/ 2bf

9/18/2013 11Lt Col A K Nigam, ITM University, Gurgaon

Page 12: Digital Modulation Unit 3

FREQUENCY‐SHIFT KEYING

• FSK is a form of constant‐amplitude anglemodulation similar to standard frequencyq ymodulation (FM) except the modulating signal isa binary signaly g

• FSK is sometimes called binary FSK (BFSK).

• The general expression for FSK is• The general expression for FSK is

9/18/2013 12Lt Col A K Nigam, ITM University, Gurgaon

Page 13: Digital Modulation Unit 3

• The modulating signal is a normalized binary waveformwhere a logic 1 = + 1 V and a logic 0 = ‐1 V.g g

• Thus, for a logic l input, we can write

F l i 0 i ( ) 1• For a logic 0 input, vm(t) = ‐1,

• Thus With binary FSK, the carrier center frequency (fc) isshifted (deviated) up and down in the frequency domain bythe binary input signal

9/18/2013 13Lt Col A K Nigam, ITM University, Gurgaon

Page 14: Digital Modulation Unit 3

• As the binary input signal changes from a logic 0 to a logic 1and vice versa, the output frequency shifts between twofrequencies

(a) mark, or logic 1 frequency (fm), (b) space, or logic 0 frequency (fs). 

*The mark and space frequencies are separated from the carrier frequency by the peak frequency ∆f deviation and from each other by 2∆f .

9/18/2013 14Lt Col A K Nigam, ITM University, Gurgaon

Page 15: Digital Modulation Unit 3

• Frequency deviation is expressed mathematically as

9/18/2013 15Lt Col A K Nigam, ITM University, Gurgaon

Page 16: Digital Modulation Unit 3

FSK Bit Rate, Baud, and Bandwidth

• The bit time equals the time of an FSK signaling element,and the bit rate equals the baud.

Th b d f bi FSK l b d t i d b• The baud for binary FSK can also be determined bysubstituting N = 1 in

• The minimum bandwidth for FSK is given as

9/18/2013 16Lt Col A K Nigam, ITM University, Gurgaon

Page 17: Digital Modulation Unit 3

Example 2‐2Determine (a) the peak frequency deviation, (b) minimumbandwidth and (c) baud for a binary FSK signal with a markbandwidth, and (c) baud for a binary FSK signal with a markfrequency of 49 kHz, a space frequency of 51 kHz, and an input bit rate of 2 kbps

9/18/2013 17Lt Col A K Nigam, ITM University, Gurgaon

Page 18: Digital Modulation Unit 3

Modulation index

• h = FM modulation index called the h‐factor in FSK

f f d t l f f th bi d l ti i l• fo = fundamental frequency of the binary modulating signal 

• ∆f = peak frequency deviation (hertz)

9/18/2013 18Lt Col A K Nigam, ITM University, Gurgaon

Page 19: Digital Modulation Unit 3

FSK Transmitter

9/18/2013 19Lt Col A K Nigam, ITM University, Gurgaon

Page 20: Digital Modulation Unit 3

FSK Receiver

(a) Noncoherent FSK demodulator

9/18/2013 20Lt Col A K Nigam, ITM University, Gurgaon

Page 21: Digital Modulation Unit 3

(b) Coherent FSK demodulator

PLL‐FSK demodulator

9/18/2013 21Lt Col A K Nigam, ITM University, Gurgaon

Page 22: Digital Modulation Unit 3

PLL‐FSK demodulator operation

• As the input to the PLL shifts between the mark and spacep pfrequencies, the dc error voltage at the output of the phasecomparator follows the frequency shift.

B th l t i t f i ( k d• Because there are only two input frequencies (mark andspace), there are also only two output error voltages.

• One represents a logic 1 and the other a logic 0.p g g

9/18/2013 22Lt Col A K Nigam, ITM University, Gurgaon

Page 23: Digital Modulation Unit 3

PerformancePerformance

• Binary FSK has a poorer error performance than PSK or QAMy p p Qand, consequently, is seldom used for high‐performancedigital radio systems.

• Its use is restricted to low‐performance, low‐cost,asynchronous data modems that are used for dataycommunications over analog, voice‐band telephone lines.

9/18/2013 23Lt Col A K Nigam, ITM University, Gurgaon

Page 24: Digital Modulation Unit 3

Continuous‐Phase Frequency‐Shift Keying

• Continuous‐phase frequency‐shift keying (CP‐FSK) is binaryFSK except the mark and space frequencies are synchronizedp p q ywith the input binary bit rate.

• This ensures a smooth phase transition in the analog outputsignal when it changes from a mark to a space frequency orvice versa.

• This has effect of limiting the BW after modulation

9/18/2013 24Lt Col A K Nigam, ITM University, Gurgaon

Page 25: Digital Modulation Unit 3

Non/continuous phase FSK waveformswaveforms

9/18/2013 25Lt Col A K Nigam, ITM University, Gurgaon

Page 26: Digital Modulation Unit 3

Comparison FSK/CP‐FSKComparison FSK/CP FSK

• CP‐FSK has a better bit‐error performance andCP FSK has a better bit error performance andlower spectral width than conventional binary FSKfor a given signal‐to‐noise ratio.

• The disadvantage of CP‐FSK is that it requiresg qsynchronization circuits and is, therefore, moreexpensive to implement.

9/18/2013 26Lt Col A K Nigam, ITM University, Gurgaon

Page 27: Digital Modulation Unit 3

PHASE‐SHIFT KEYING• The simplest form of PSK is binary phase‐shift keyingp y p y g(BPSK), where N = 1 and M = 2.

• Therefore, with BPSK, two phases (2^1 = 2) areibl f h ipossible for the carrier.

• One phase represents a logic 1, and the other phaserepresents a logic 0represents a logic 0.

• As the input digital signal changes state (i.e., from a 1to a 0 or from a 0 to a 1), the phase of the output), p pcarrier shifts between two angles that are separated by180°.

h f h l k i ( )• Other names for BPSK are phase reversal keying (PRK)and bi‐phase modulation.BPSK i f f d l ti f• BPSK is a form of square‐wave modulation of acontinuous wave (CW) signal.

9/18/2013 27Lt Col A K Nigam, ITM University, Gurgaon

Page 28: Digital Modulation Unit 3

BPSK ModulatorBPSK Modulator

9/18/2013 28Lt Col A K Nigam, ITM University, Gurgaon

Page 29: Digital Modulation Unit 3

Balanced Modulator

9/18/2013 29Lt Col A K Nigam, ITM University, Gurgaon

Page 30: Digital Modulation Unit 3

BPSK modulator: truth table phasor and  constellation diagram

9/18/2013 30Lt Col A K Nigam, ITM University, Gurgaon

Page 31: Digital Modulation Unit 3

Bandwidth considerations of BPSK

.cos (assuming unit amplitude)m cout put cos w t w t=

( )1 [cos( ) cos( )]

neglecting higher frequencies

w w t w w= + + −[cos( ) cos( )]2

2 putting this in aboveeq weget

c m c m

b

w w t w w

fBut here w π

= + +

= putting this in aboveeq.weget2

1 [cos( ) cos( )]

m

b b

But here w

f ff t f

=

= + +[cos( ) cos( )]2 2 2

( ) ( )

c c

b b

f t f

f fBW f f f

= + + −

+( ) ( )2 2b b

c c bBW f f f= + − − =

9/18/2013 31Lt Col A K Nigam, ITM University, Gurgaon

Page 32: Digital Modulation Unit 3

Output phase‐versus‐time relationship for a BPSKmodulator

9/18/2013 32Lt Col A K Nigam, ITM University, Gurgaon

Page 33: Digital Modulation Unit 3

BPSK receiver.

9/18/2013 33Lt Col A K Nigam, ITM University, Gurgaon

Page 34: Digital Modulation Unit 3

Demodulation

Similarly it can be done for logic 0 (do your self9/18/2013 34Lt Col A K Nigam, ITM University, Gurgaon

Page 35: Digital Modulation Unit 3

Quaternary Phase‐Shift KeyingQuaternary Phase Shift Keying

• QPSK is an M‐ary encoding scheme where N = 2 andQPSK is an M ary encoding scheme where N   2 and M= 4

• Therefore, with QPSK, the binary input data are , Q , y pcombined into groups of two bits, called dibits.

• each dibit code generates one of the four possible g poutput phases (+45°, +135°, ‐45°, and ‐135°).

9/18/2013 35Lt Col A K Nigam, ITM University, Gurgaon

Page 36: Digital Modulation Unit 3

QPSK transmitter.

9/18/2013 36Lt Col A K Nigam, ITM University, Gurgaon

Page 37: Digital Modulation Unit 3

QPSK transmitter.

• Two bits (di‐bit) are clocked into the bit splitter.

• After both bits have been serially inputted, they areAfter both bits have been serially inputted, they aresimultaneously parallel outputted.

• The I bit modulates a carrier that is in phase withpthe reference oscillator (hence the name "I" for "inphase" channel)

• The Q bit modulate, a carrier that is 90° out ofphase i.e. cosine wave .

9/18/2013 37Lt Col A K Nigam, ITM University, Gurgaon

Page 38: Digital Modulation Unit 3

l i d l i h ibl h• For a logic 1 = + 1 and a logic 0= ‐ 1 , two phases are possible at theoutput of the –I modulator (+sinwct and ‐ sinwct)

Si il l t h ibl t th t t f th Q b l d• Similarly two phases are possible at the output of the Q balancedmodulator (+coswct), and (‐coswct).

• For input of Q =I= 1 the two inputs to the I balanced modulator are• For input of Q =I= 1, the two inputs to the I balanced modulator are +1 and sinwct, and The two inputs to the Q balanced modulator are +1 and coswct.

• Outputs are(a) I balanced modulator =(+1)(sinwct) = +1 sinwct(b) Q balanced modulator =(+1)(coswct) = +1 coswct(b) Q balanced modulator  (+1)(coswct)   +1 coswct

9/18/2013 38Lt Col A K Nigam, ITM University, Gurgaon

Page 39: Digital Modulation Unit 3

Output of the linear summer for 1,1 input dibits is

cosc csin w t w t= +

{ }0

0 0

(90 )

2 90 90

c csin w t sin w t

t

= + +

⎧ ⎫⎛ ⎞ ⎛ ⎞⎪ ⎪0 02 90 902sin cos2 2

cw t⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪= ⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠⎩ ⎭

02 902 sin2

cw t⎛ ⎞+= ⎜ ⎟

⎝ ⎠2

1.414sin( 45 }ocw t

⎝ ⎠= +

(Similarly it can be calculated for all other input combinations)9/18/2013 39Lt Col A K Nigam, ITM University, Gurgaon

Page 40: Digital Modulation Unit 3

Truth tableTruth table

9/18/2013 40Lt Col A K Nigam, ITM University, Gurgaon

Page 41: Digital Modulation Unit 3

Phasor diagram

9/18/2013 41Lt Col A K Nigam, ITM University, Gurgaon

Page 42: Digital Modulation Unit 3

Constellation diagram

(How to remember: For 1 +ve will go and for 0 –ve will go)I is for sine and Q for cosine

9/18/2013 42Lt Col A K Nigam, ITM University, Gurgaon

Page 43: Digital Modulation Unit 3

Salient featuresSalient features

• Each of the four possible output phasors has exactly theac o t e ou poss b e output p aso s as e act y t esame amplitude. Therefore, the binary informationmust be encoded entirely in the phase of the outputi lsignal.

• The angular separation between any two adjacentphasors in QPSK is 90°phasors in QPSK is 90 .

• Thus a QPSK signal can undergo almost a +45° or ‐45°shift in phase during transmission and still retain thep gcorrect encoded information when demodulated at thereceiver.

9/18/2013 43Lt Col A K Nigam, ITM University, Gurgaon

Page 44: Digital Modulation Unit 3

Output phase‐versus‐time relationship for a QPSKmodulatormodulator.

9/18/2013 44Lt Col A K Nigam, ITM University, Gurgaon

Page 45: Digital Modulation Unit 3

Bandwidth considerations of QPSK

The highest fundamental frequency at the input and fastest rate of change at the output of the balance modulators is equal to one‐fourth of the binary input bit rate.

9/18/2013 45Lt Col A K Nigam, ITM University, Gurgaon

Page 46: Digital Modulation Unit 3

Mathematical Analysis

.cos (assuming unit amplitude)( )

m cout put cos w t w tneglecting higher frequencies=

1 [cos( ) cos( )]2 c m c mw w t w w= + + −

2 putting this in aboveeq.weget4

bm

fBut here w π=

1 [cos( ) cos( )]2 4 4

b bc c

f ff t f= + + −

( ) ( )4 4 2b b b

c cf f fBW f f= + − − =

9/18/2013 46Lt Col A K Nigam, ITM University, Gurgaon

Page 47: Digital Modulation Unit 3

QPSK receiver

9/18/2013 47Lt Col A K Nigam, ITM University, Gurgaon

Page 48: Digital Modulation Unit 3

Output of the I product detectorOutput of the I product detector

(For input I=0 and Q=1)

9/18/2013 48Lt Col A K Nigam, ITM University, Gurgaon

Page 49: Digital Modulation Unit 3

Output of the Q product detector 

9/18/2013 49Lt Col A K Nigam, ITM University, Gurgaon

Page 50: Digital Modulation Unit 3

Offset QPSK.I difi d f f QPSK h h bi h I d Q• Is a modified form of QPSK where the bits on the I and Qchannels are offset in phase from each other by one‐half of a bittime.

• Because changes in the i channel occur at the midpoints of the qchannel bits and vice versa, there is never more than a single bitchange in the dibitchange in the dibit

• Therefore, there is never more than a 90° shift in the outputphase compared to conventional QPSK in which a change in thep p ginput dibit from 00 to 11 or 01 to 10 causes a corresponding180° shift in the output phase.

Ad t f OQPSK i th li it d h hift th t t b• Advantage of OQPSK is the limited phase shift that must beimparted during modulation.

• disadvantage of OQPSK is that changes in the output phase g Q g p poccur at twice the data rate in either the I or Q channel".

9/18/2013 50Lt Col A K Nigam, ITM University, Gurgaon

Page 51: Digital Modulation Unit 3

Offset keyed (OQPSK)

9/18/2013 51Lt Col A K Nigam, ITM University, Gurgaon

Page 52: Digital Modulation Unit 3

Baud and minimum bandwidthBaud and minimum bandwidth

• OQPSK the baud and minimum bandwidthOQPSK the baud and minimum bandwidth are twice that of conventional QPSK for a given transmission bit rategiven transmission bit rate. 

• OQPSK is sometimes called OKQPSK (offset‐keyed QPSK)keyed QPSK).

9/18/2013 52Lt Col A K Nigam, ITM University, Gurgaon

Page 53: Digital Modulation Unit 3

8‐PSK Modulation• With 8 PSK three bits are encoded forming tribits and• With 8‐PSK, three bits are encoded, forming tribits and

producing eight different output phases.

• To encode eight different phases, the incoming bits areencoded in groups of three, called tribits (2^3 = 8)

9/18/2013 53Lt Col A K Nigam, ITM University, Gurgaon

Page 54: Digital Modulation Unit 3

8‐PSK transmitter.

How to find anglesFor 111 inputFor 111 inputOutput= 1.307sinwt+0.541 coswtAngle= tan^‐1(.541/1.307) in 1st quadrangle=67.5°9/18/2013 54Lt Col A K Nigam, ITM University, Gurgaon

Page 55: Digital Modulation Unit 3

Output Phases

NotePhases are +/ (22 5°+45°)Phases are +/‐(22.5°+45°)

9/18/2013 55Lt Col A K Nigam, ITM University, Gurgaon

Page 56: Digital Modulation Unit 3

Output phase‐versus‐time relationship for an 8‐PSK modulatorPSK modulator

9/18/2013 56Lt Col A K Nigam, ITM University, Gurgaon

Page 57: Digital Modulation Unit 3

Phasor DiagramPhasor Diagram

9/18/2013 57Lt Col A K Nigam, ITM University, Gurgaon

Page 58: Digital Modulation Unit 3

Constellation Diagram: 8 PSK

Note: Format is QIC

I    ‐ve (0)                                                                              I   +ve (1)Q  +ve (1) Q  +ve (1)Q ve ( ) Q ve ( )

I    ‐ve (0) I     +veQ  +ve (1) Q   ‐veQ ( ) Q

9/18/2013 58Lt Col A K Nigam, ITM University, Gurgaon

Page 59: Digital Modulation Unit 3

Bandwidth considerations of 8‐PSK• With 8‐PSK, the data are divided into three channels,• the bit rate in the I, Q, or C channel is equal to one‐third of the binary input 

data rate (fb /3).

A l iAnalysiscos . ( )

( )c mout put w t cos w t amplitudeassumed tobeunity

neglecting higher frequencies=( )

1 [cos ( ) cos ( )]2 c m c m

g g g f q

w w t w w= + + −

2 putting this in aboveeq.weget6

bm

fBut here w π=

1 [cos ( ) cos ( )]2 6 6

b bc c

f ff t f

f f f

= + + −

( ) ( )6 6 3b b b

c cf f fBW f t f= + − − =

9/18/2013 59Lt Col A K Nigam, ITM University, Gurgaon

Page 60: Digital Modulation Unit 3

8‐PSK receiver.

9/18/2013 60Lt Col A K Nigam, ITM University, Gurgaon

Page 61: Digital Modulation Unit 3

8 PSK Performance8 PSK Performance

• With 8‐PSK, the angular separation between adjacent output , g p j pphases is only 45° (360 / 8 ). 

• Therefore, 8‐PSK can undergo only a 22.5° phase shift during transmission and still retain its integrity.

9/18/2013 61Lt Col A K Nigam, ITM University, Gurgaon

Page 62: Digital Modulation Unit 3

16‐PSK

Truth table

9/18/2013 62Lt Col A K Nigam, ITM University, Gurgaon

Page 63: Digital Modulation Unit 3

Constellation diagram

9/18/2013 63Lt Col A K Nigam, ITM University, Gurgaon

Page 64: Digital Modulation Unit 3

16 PSK Performance16 PSK Performance

• With 16‐PSK, the angular separation between adjacent , g p joutput phases is only 22.5° (360 / 16). 

• Therefore, 16‐PSK can undergo only a 11.25° phase shift during transmission and still retain its integrity.

9/18/2013 64Lt Col A K Nigam, ITM University, Gurgaon

Page 65: Digital Modulation Unit 3

QUADRATURE – AMPLITUDE MODULATION

8‐QAM

• 8‐QAM is an M‐ary encoding techniquewhere M = 8.

• Unlike 8‐PSK, the output signal from an 8‐QAM modulator is not a constant‐amplitudesignal.

9/18/2013 65Lt Col A K Nigam, ITM University, Gurgaon

Page 66: Digital Modulation Unit 3

8‐QAM modulator 

9/18/2013 66Lt Col A K Nigam, ITM University, Gurgaon

Page 67: Digital Modulation Unit 3

• The incoming data are divided into groups of three bits (tribits): the

8‐QAM modulator • The incoming data are divided into groups of three bits (tribits): the

I, Q, and C bit streams.

E h t h bit t l t thi d f th i i d t• Each stream has a bit rate equal to one‐third of the incoming datarate.

• The I and Q bits determine the polarity of the PAM signal at theoutput of the 2‐to‐4‐level converters

• The C channel determines the magnitude.

• Because the c bit is fed un‐inverted to both the i and the q channelBecause the c bit is fed un inverted to both the i and the q channel2‐to‐4‐level converters, the magnitudes of the I and Q PAM signalsare always equal.

• Their polarities depend on the logic condition of the i and q bitsand, therefore, may be different.9/18/2013 67Lt Col A K Nigam, ITM University, Gurgaon

Page 68: Digital Modulation Unit 3

Truth TableTruth Table

9/18/2013 68Lt Col A K Nigam, ITM University, Gurgaon

Page 69: Digital Modulation Unit 3

Phasor and constellation diagramg

9/18/2013 69Lt Col A K Nigam, ITM University, Gurgaon

Page 70: Digital Modulation Unit 3

Output phase and amplitude‐versus‐timerelationship for 8 QAMrelationship for 8‐QAM

9/18/2013 70Lt Col A K Nigam, ITM University, Gurgaon

Page 71: Digital Modulation Unit 3

Bandwidth considerations of 8‐QAM.

• N=3N=3

• Thus the minimum bandwidth required for 8‐QAM is fb / 3 the same as in 8 PSKQAM is fb / 3, the same as in 8‐PSK.

9/18/2013 71Lt Col A K Nigam, ITM University, Gurgaon

Page 72: Digital Modulation Unit 3

8‐QAM receiver.

• An 8‐QAM receiver is almost identical to theAn 8 QAM receiver is almost identical to the 8‐PSK receiver

9/18/2013 72Lt Col A K Nigam, ITM University, Gurgaon

Page 73: Digital Modulation Unit 3

16‐QAM

• As with the 16‐PSK 16‐QAM is an M‐aryAs with the 16 PSK, 16 QAM is an M arysystem where M =16.

• The input data are acted on in groups of four• The input data are acted on in groups of four(2^4 = 16).

A i h 8 QAM b h h h d h• As with 8‐QAM, both the phase and theamplitude of the transmit carrier are varied.

9/18/2013 73Lt Col A K Nigam, ITM University, Gurgaon

Page 74: Digital Modulation Unit 3

QAM transmitter

• The input binary data are divided into four channels: I, I', Q, and Q'.

• The bit rate in each channel is equal to one‐fourth of the input bit rateThe bit rate in each channel is equal to one fourth of the input bit rate (fb/4).

• The I and Q bits determine the polarity at the output of the 2 to 4 level• The I and Q bits determine the polarity at the output of the 2‐to‐4‐level converters

• logic 1 = positive• logic 0 =negative

• The I' and Q' bits determine the magnitudeThe I  and Q  bits determine the magnitude • logic 1 = 0.821V • logic 0 = 0.22 V

9/18/2013 74Lt Col A K Nigam, ITM University, Gurgaon

Page 75: Digital Modulation Unit 3

9/18/2013 75Lt Col A K Nigam, ITM University, Gurgaon

Page 76: Digital Modulation Unit 3

QAMtransmitter

I and Q, 1 = +ve logic 0 =‐ve

I’ and Q’ logic 1 = 0.821V, logic 0 = 0.22 V

9/18/2013 76Lt Col A K Nigam, ITM University, Gurgaon

Page 77: Digital Modulation Unit 3

Computation of values of amplitude and phasesComputation of values of amplitude and phases

• The outputs from the I and Q channel product modulators arecombined in the linear summer and produce a modulated output

• For a quadbit input of I= 1, I' = 0, Q = 1, and Q' = 0 i.e.(1010) logic 1q p , , Q , Q ( ) gfor I and Q gives +ve value and logic 0 for I’ and Q’ gives 0.22 V

•• Thus summer output = 0.22 sin wc t + 0.22cos wc tThus summer output 0.22 sin wc t + 0.22cos wc t

=0.22{sinwct + sin(90 + wct)}=0.22[2{sin(wct + 45). cos45}=0.311sin(wct+45)

• Similarly other values of amplitude and phase can be computed

9/18/2013 77Lt Col A K Nigam, ITM University, Gurgaon

Page 78: Digital Modulation Unit 3

Values of amplitude and phases

9/18/2013 78Lt Col A K Nigam, ITM University, Gurgaon

Page 79: Digital Modulation Unit 3

Phasor and constellation diagramg

1 0 2 3

3

1          0                    2          3

2

0

1

How to remember9/18/2013 79Lt Col A K Nigam, ITM University, Gurgaon

Page 80: Digital Modulation Unit 3

Bandwidth considerations of 16‐QAM.

9/18/2013 80Lt Col A K Nigam, ITM University, Gurgaon

Page 81: Digital Modulation Unit 3

Bandwidth considerations of 16‐QAM.

9/18/2013 81Lt Col A K Nigam, ITM University, Gurgaon

Page 82: Digital Modulation Unit 3

BANDWIDTH EFFICIENCY

• Bandwidth efficiency isBandwidth efficiency is

9/18/2013 82Lt Col A K Nigam, ITM University, Gurgaon

Page 83: Digital Modulation Unit 3

9/18/2013 83Lt Col A K Nigam, ITM University, Gurgaon

Page 84: Digital Modulation Unit 3

Example

Qu.  For 16‐PSK and a transmission system with a 10 kHz bandwidth, determine the maximum ,bit rate.

Solutionh b d id h ffi i f i hi h• The bandwidth efficiency for 16‐PSK is 4, which means that four bits can be propagated through the system for each hertz of bandwidth.the system for each hert of bandwidth.

• Therefore, the maximum bit rate is simply the product of the bandwidth and the bandwidth ffi iefficiency, or

• bit rate = 4 x 10,000 = 40,000 bps

9/18/2013 84Lt Col A K Nigam, ITM University, Gurgaon

Page 85: Digital Modulation Unit 3

9/18/2013 85Lt Col A K Nigam, ITM University, Gurgaon

Page 86: Digital Modulation Unit 3

DIFFERENTIAL PHASE‐SHIFT KEYING

Is an alternative form of digitalgmodulation where the binary inputinformation is contained in the differencebetween two successive signalingelements rather than the absolute phase.p

9/18/2013 86Lt Col A K Nigam, ITM University, Gurgaon

Page 87: Digital Modulation Unit 3

OP of mod is same for input as ‘1’ and is shifted by 180° for input ‘0’

Input Data

(initial reference bit is assumed a logic 0, If the initialreference bit is assumed a logic 1, the output from theXNOR circuit is simply the complement of that shown)

9/18/2013 87Lt Col A K Nigam, ITM University, Gurgaon

Page 88: Digital Modulation Unit 3

9/18/2013 88Lt Col A K Nigam, ITM University, Gurgaon

Page 89: Digital Modulation Unit 3

DemodulationDemodulation

9/18/2013 89Lt Col A K Nigam, ITM University, Gurgaon

Page 90: Digital Modulation Unit 3

DemodulationDemodulation

Change of phase indicates 0 same phase indicates 1Change of phase indicates 0, same phase indicates 1

9/18/2013 90Lt Col A K Nigam, ITM University, Gurgaon

Page 91: Digital Modulation Unit 3

PROBABILITY OF ERROR AND BIT ERROR RATE• Probability of error P(e) and bit error rate (BER) are often used• Probability of error P(e) and bit error rate (BER) are often used

interchangeably

• It is a function of the carrier‐to‐noise power ratio (or, morespecifically, the average energy per bit‐to‐noise power densityratio) and the number of possible encoding conditions used (M‐ary)ary).

• Energy per bit is simply the energy of a single bit of information.Mathematically,

Energy per bit is Eb = C.Tb=C/fb.............................1

• Noise power density is the thermal noise power normalized to a1 H b d idth (i th i t i 1 H1‐ Hz bandwidth (i.e., the noise power present in a 1‐Hzbandwidth). Mathematically, noise power density is

No = N/B N= Noise, B= BW…………………………2/ ,

9/18/2013 91Lt Col A K Nigam, ITM University, Gurgaon

Page 92: Digital Modulation Unit 3

From eq 1 and 2 we can writeFrom eq 1 and 2 we can write

//

b bE C fN N B

=0 /

B

N N B

C BN f

⎛ ⎞⎛ ⎞= ×⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠B

b

N fStated in dB

E C B

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎛ ⎞ +⎜ ⎟ ⎜ ⎟⎜ ⎟0

b

dB B dBdBN N f

or

⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

0

b

dBB dB dB

EB Cf N N

⎛ ⎞⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

9/18/2013 92Lt Col A K Nigam, ITM University, Gurgaon

Page 93: Digital Modulation Unit 3

PSK Error Performance• For PSK systems, the phase

difference between two consecutivei li i t i 2 / Msignaling points is α =2π / M

• For error free transmission themaximum shift of angle permissible α =2 π/2Mg pis α= 2π /2M either way fromsignaling point…………………………….1

If d di b i

α =2 π/2Md

D

• If d= distance between consecutivetwo signaling points and D=Peaksignal amplitude then

sin α=(d/2)/D or

• Sin(2π /2M)=d/2D from Eq. 1

or d/2 =D.sin(π/M)…………………….2

9/18/2013 93Lt Col A K Nigam, ITM University, Gurgaon

Page 94: Digital Modulation Unit 3

Computation of Bit Error Probability

• If no. of states =M, then no. of bits required to encode these=log2M

If E /N i th bit t i d it ti th i• If Eb/N0 is the bit energy to noise density ratio then energy in all bits

log bED M⎛ ⎞

= × ⎜ ⎟

• Putting this in eq. 2 we get

20

logD MN

= × ⎜ ⎟⎝ ⎠

20

/ 2 log ................3bEd Sin MM Nπ ⎛ ⎞⎛ ⎞= × ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠• Now the probability that the noise will be greater than z is 

given by error function

2 ∞22( ) z

z

P z e dzπ

∞−= ∫

9/18/2013 94Lt Col A K Nigam, ITM University, Gurgaon

Page 95: Digital Modulation Unit 3

• Now maximum error can be d/2, thus error will be obtained / ,by putting z=d/2 from eq. 3

• As there are log2M bits per signal, the bit error probability ld bwould be

1( ) ( )log

P e erfc zM

=

• Where Z is given by

2log M

g y

20

/ 2 log ................3bEz d Sin MM Nπ ⎛ ⎞⎛ ⎞= = × ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

• And M is no. of levels

9/18/2013 95Lt Col A K Nigam, ITM University, Gurgaon

Page 96: Digital Modulation Unit 3

Example of BPSKExample of BPSK

M=2, thus from eq. 3 we get

20

/ 2 log 22

bEd SinN

E

π ⎛ ⎞⎛ ⎞= × ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

0

bEN

=

1( )l 2

b bE EP e erfc erfcN N

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2 0 0log 2 N N⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

9/18/2013 96Lt Col A K Nigam, ITM University, Gurgaon

Page 97: Digital Modulation Unit 3

Example of QPSKExample of QPSK

/ 2 log 4 bEd Si π ⎛ ⎞⎛ ⎞ × ⎜ ⎟⎜ ⎟ 20

/ 2 log 44

1 2

b

b b

d SinN

E E

= × ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟

0 0

22 N N⎜ ⎟

⎝ ⎠

2 0 0

1( )log 2

b bE EP e erfc erfcN N

= =2 0 0g

9/18/2013 97Lt Col A K Nigam, ITM University, Gurgaon

Page 98: Digital Modulation Unit 3

Error rates ofError rates of PSK modulationmodulation systems

9/18/2013 98Lt Col A K Nigam, ITM University, Gurgaon

Page 99: Digital Modulation Unit 3

QAM Error Performance

1 1( ) ( )l

LP e erfc zL L

−⎛ ⎞= ⎜ ⎟⎝ ⎠2

( ) ( )log

fL L

Where z is givenby

⎜ ⎟⎝ ⎠

2

0

log1

bL EzL N

= ×− 0

9/18/2013 99Lt Col A K Nigam, ITM University, Gurgaon

Page 100: Digital Modulation Unit 3

FSK Error Performance

• The probability of error for coherent FSK is

1( )2 2

bEP e erfcN

=02 2N

9/18/2013 100Lt Col A K Nigam, ITM University, Gurgaon

Page 101: Digital Modulation Unit 3

ASK Error Performance

• The probability of error for coherent ASK isp y

1( ) bEP e erfc⎛ ⎞

= ⎜ ⎟⎜ ⎟0

( )2 2

P e erfcN⎜ ⎟⎜ ⎟

⎝ ⎠

• The probability of error for non‐coherent FSK is

1 E⎛ ⎞

0

1( )2 2

bEP e erfcN

⎛ ⎞−= ⎜ ⎟

⎝ ⎠

9/18/2013 101Lt Col A K Nigam, ITM University, Gurgaon

Page 102: Digital Modulation Unit 3

DPSK Error PerformanceDPSK Error Performance

01( )

bENP e e−

=( )2

P e e=

9/18/2013 102Lt Col A K Nigam, ITM University, Gurgaon

Page 103: Digital Modulation Unit 3

Finding Error function values usingFinding Error function values using approximate formula

2

( )xef

( )erfc xx π

=

For exampleperfc(.7)=.493erfc(.5)=.878

9/18/2013 103Lt Col A K Nigam, ITM University, Gurgaon

Page 104: Digital Modulation Unit 3

BANDWIDTH EFFICIENCY

• Also called information density or spectral efficiency, often y p y,used to compare the performance of one digital modulation technique to another.

B d idth ffi i i d fi d• Bandwidth efficiency is defined as

9/18/2013 104Lt Col A K Nigam, ITM University, Gurgaon

Page 105: Digital Modulation Unit 3

Clock RecoveryClock Recovery

• Squaring LoopSquaring Loop

• Costas Loop

d l• Remodulator

9/18/2013 105Lt Col A K Nigam, ITM University, Gurgaon

Page 106: Digital Modulation Unit 3

Squaring Loop• The incoming modulated signal is squared and band‐pass

filtered to extract the carrier component at 2 times itsoriginal frequencyoriginal frequency.

• This signal is then fed into a phase locked loop whose otherinput comes from a VCO.

• The error output of the phase locked loop is converted into aDC voltage which is fed back into the VCO to cause it tooscillate at a frequency which is almost same as the carrieroscillate at a frequency which is almost same as the carrierfrequency such that the error output reduces to nearly zero.

• This is then divided by two to give the in phase carrierfrequency

9/18/2013 106Lt Col A K Nigam, ITM University, Gurgaon

Page 107: Digital Modulation Unit 3

BPF DividerPLLSquarer

i ( )I t t t PSK i t±

BPF DividerPLLSquarer

2

sin ( )

i

cInput to squarer w t PSK inputOutput of squarer

= ±

2sin1 (1 cos 2 )2

c

c

w t

w t

= +

= + −( )2

1

c

This is filtered to give1 cos 22

2

cw t

This is divided by to givein phasecarrier frequency

= −

2This is divided by to givein phasecarrier frequency

9/18/2013 107Lt Col A K Nigam, ITM University, Gurgaon

Page 108: Digital Modulation Unit 3

COSTAS LOOP

9/18/2013 108Lt Col A K Nigam, ITM University, Gurgaon

Page 109: Digital Modulation Unit 3

• It consists of two coherent detectors supplied with the samereceived PSK inputreceived PSK input

• Carrier is generated locally by using a VCO which is having aphase difference of φ for simplicity we assume that it hasamplitude =1 volt.

• This carrier is given as it is to I product modulator

O h d fi i li d h l ll d i• Other product figure is applied the locally generated carrierwith phase of 90° shift of as shown.

• Both the outputs of I and Q channel are passed through aBoth the outputs of I and Q channel are passed through aLPF and are fed to a phase discriminator which is consistingof a multiplier followed by a low pass filter

f h f l l h h l• Output of the final LPF is error voltage which is proportionalto sin2φ and it corrects VCO frequency to carrier frequencyby bringing to φ to 0y g g φ

9/18/2013 109Lt Col A K Nigam, ITM University, Gurgaon

Page 110: Digital Modulation Unit 3

. c o s . ( ) c o s ( )c c c

O u tp u t o f I M o d u la to r isA w t m t w tA

φ= × +

{ }( ) c o s ( 2 ) c o s2

( ) c o s2

cc

c

A m t w t

AA fte r L P F it is m t

φ φ

φ

= + +

= → →2

. c o s . ( ) s in ( )c c c

O u tp u t o f Q M o d u la to r isA w t m t w t φ= × +

{ }

{ }

( ) s in ( 2 ) s in ( )2

( ) s in ( 2 ) s in

cc

c

A m t w t

A m t w t

φ φ

φ φ

= + − −

= + +{ }( ) s in ( 2 ) s in2 cm t w t

A f te r L P

φ φ+ +

( ) s in2

cAF i t is m t φ=

22

m in1( ) c o s . ( ) s in s in 2 ( )

2 2 4 2c c c

O u tp u t o f p h a s e d is c r i a to r isA A Am t m t m tφ φ φ⎡ ⎤= = ⎢ ⎥⎣ ⎦

0 0a s e r r o r v o l ta g e g o e s toa n d i t lo c k s to th e c a r r ie r fr e q u e n c y

φ⎣ ⎦

→9/18/2013 110Lt Col A K Nigam, ITM University, Gurgaon

Page 111: Digital Modulation Unit 3

Clock recovery and timing diagramClock recovery and timing diagram

9/18/2013 111Lt Col A K Nigam, ITM University, Gurgaon